C++: &a[2] - &a[1] ==?

Question

a is array of integers, if I try to subtract the address value of &a[2] - &a[1] == ? what should the result be 4 or 1 ?

EDIT: see 4th comment on top answer here why he says 1 ?? this is why I'm confused I thought it will be 4

EDIT: here is a test

@disioe : you don't have a compiler? So why are you writing code? — Mitch Wheat, May 24 '11 at 05:40
@Mitch Wheat: i'm not I just can't sleep, so I'm reading stackoverflow:)) I program in C# for money. But actually I was going to try it on codepad.org but it's not working now so I decided to ask — disioe, May 24 '11 at 05:41
@disioe: I'm sorry, I have no idea what you are talking about? — Mitch Wheat, May 24 '11 at 06:25

Mihran Hovsepyan · Accepted Answer · 2013-07-01T08:22:01.870

9

&a[2] is same as &(*(a + 2)) (i.e (a + 2)) and &a[1] is same as &(*(a + 1)) (i.e. (a + 1)). So answer will be 1.

edited Jul 01 '13 at 08:22

answered May 24 '11 at 05:38

Mihran Hovsepyan

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Thank you very much, I get it now. It was in the thread I read too but somehow I didn't get it. Thanks again for your help. – disioe May 24 '11 at 05:48

score 6 · Answer 2 · answered May 24 '11 at 05:38

6

Pointer subtraction gives you the difference in elements, not bytes. It does not matter what the element type of the array is, the result of &a[2] - &a[1] will always be 1, because they are 1 element apart.

answered May 24 '11 at 05:38

fredoverflow

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score 4 · Answer 3 · answered May 24 '11 at 05:39

It is always 1. The pointer arithmetics are not concerned with the number of bytes that each element has, and this is very useful. Compare these:

ptr++;  // go to the next element (correct)
ptr += sizeof *ptr;  // go to the next element (wrong)

When you work with arrays you are usually interested in the elements, not in the bytes comprising them, and that is why pointer arithmetics in C has been defined this way.

score 3 · Answer 4 · answered May 24 '11 at 05:38

3

The difference must be 1. When you compare pointers you will always get the difference of elements.

answered May 24 '11 at 05:38

harper

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score 2 · Answer 5 · answered May 24 '11 at 05:47

2

Since this is C++, I'm going to assume that you have not overridden the & or * operators on whatever type a is. Minding that, the following is true:

&a[2] - &a[1]
(a + 2) - (a + 1)
a + 2 - a - 1
2 - 1
1

answered May 24 '11 at 05:47

Bill Lynch

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sean e · Answer 6 · 2011-05-24T14:26:20.640

1

A couple of the answers here (deleted since this answer was posted) clearly had byte* in mind:

    int a[10];
    byte * pA2 = (byte*)&a[2];
    byte * pA1 = (byte*)&a[1];
    int sz1 = &a[2] - &a[1];
    int sz2 = pA2 - pA1;
    CString msg;
    msg.Format("int * %d, byte * %d\n", sz1, sz2);
    OutputDebugString(msg);

output is:

     int * 1, byte * 4

Two addresses but depending on the declaration of the variable the addresses are stored in, the difference between the two can be 1 or 4.

edited May 24 '11 at 14:26

answered May 24 '11 at 05:55

sean e

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1

That was my problem, as the question was not in code, I read the question as what is the difference between two addresses, not in terms of "pointer arithmetic". – Jens May 24 '11 at 07:32

C++: &a[2] - &a[1] ==?

6 Answers6